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1. Andrei, Boris, and Valentin participated in a 1 km race. (We assume that each of them ran at a constant speed). Andrei was 50 m ahead of Boris at the finish line. And Boris was 40 m ahead of Valentin at the finish line. What was the distance between Andrei and Valentin at the moment Andrei finished?
|
Answer: 88 m.
Solution. Let the speeds of Andrey, Boris, and Valentin be $a, b$, and $c$ m/s, respectively. From the condition, it follows that $b=0.95 a, c=0.96 b$. Therefore, $c=0.96 \cdot 0.95 a=0.912 a$. This means that when Andrey runs 1000 m, Valentin will cover 912 m. The lag will be 88 m.
Evaluation. 12 points for a correct solution.
|
88
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Two identical resistors with resistance $R$ each are connected in series and connected to a source of constant voltage $U$. An ideal voltmeter is connected in parallel to one of the resistors. Its readings were $U_{v}=10 B$. After that, the voltmeter was replaced with an ideal ammeter. The ammeter readings were $-I_{A}=10 \mathrm{~A}$. Determine the value of $R$. (10 points)
|
Answer: 2 Ohms
Solution. Voltage of the source: $U=U_{v}+U_{v}=20 V$ (4 points). The resistance of an ideal ammeter: $r_{A}=0$ Ohms (3 points). Therefore, the resistance of the resistor: $R=\frac{U}{I_{A}}=\frac{20}{10}=2$ Ohms (3 points).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The structure shown in the figure is in equilibrium. It is known that the length of the homogeneous rod $l=50 \mathrm{~cm}$, and its mass $m_{2}=2$ kg. The distance between the attachment points of the left thread to the rod $S=10$ cm. Determine the mass $m_{1}$ of the load. All threads are weightless and inextensible. The pulleys are weightless. (15 points)
|
Answer: 10 kg
Solution. From the equilibrium condition for the large block, it follows that the tension force in the left thread: $T_{n}=\frac{m g}{2}$ (5 points). The equilibrium condition for the rod relative to the attachment point of the right thread: $T_{n} \cdot l=T_{n} \cdot(l-S)+m_{2} g \cdot \frac{1}{2} l$ (5 points).
As a result, we get: $m_{1}=\frac{m_{2} l}{S}=\frac{2 \cdot 0.5}{0.1}=10$ kg (5 points).
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A palindrome is a number that reads the same from left to right and from right to left. For example, the numbers 353 and $4884-$ are palindromes. It is known that a three-digit number $x$ is a palindrome. To it, 32 was added and a four-digit number was obtained, which is also a palindrome. Find $x$.
|
# Answer: 969.
Solution. Since $x+32$ is a four-digit number, the inequality $x+32 \geqslant 1000$ holds, from which $x \geqslant 968$. From the fact that $x$ is a three-digit palindrome, we get that this number starts and ends with the digit 9, and its middle digit is not less than 6. By checking the four possible options, we find the only answer.
Evaluation. If the answer is found but its uniqueness is not proven, 6 points. For a complete solution, 12 points.
|
969
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the cells of a $3 \times 3$ square, the numbers $1,2,3, \ldots, 9$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$
#
|
# Answer: 7.
Solution. Let's color the cells in a checkerboard pattern: let the corner and central cells be black, and the rest white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white ones, we get that the odd numbers are in the black cells. Their total sum is $1+3+5+7+9=$ 25. Therefore, the number in the central cell is $7=25-18$.
Evaluation. 13 points for a correct solution. If an example of the arrangement of numbers satisfying the problem's condition is provided but the uniqueness of the answer is not proven, 6 points.
## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences
## Final Stage
$2017-2018$ academic year
## Problems, answers, and evaluation criteria
## 6th Grade Variant 1
## physics
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. An aquarium in the shape of a rectangular parallelepiped has dimensions: length $1.5 m$, width - $400 mm$, height - 80 cm. It is filled with water at a rate of 2 liters/minute. How many seconds after the start of filling will the aquarium be completely filled? (10 points)
|
Answer: 14400 s
Solution. Volume of the aquarium $\quad V=1.5 \cdot 0.4 \cdot 0.8=0.48 \mathrm{~m}^{3}$.
Filling rate $v=\frac{0.002 \mu^{3}}{60 s}$ (3 points). The aquarium will be completely filled in time $\quad t=\frac{V}{v}=\frac{0.48 \cdot 60}{0.002}=14400 ~ s \quad(4$ points $)$.
|
14400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Two people are walking towards each other from points $A$ and $B$ with speeds $v_{1}=6 \mathrm{~m} / \mathrm{s}$ and $v_{2}=4 \mathrm{~m} / \mathrm{s}$. At the moment of their meeting, one of them turned around and walked in the opposite direction, while the other continued in the same direction. The person who turned around, upon returning to their starting point, waited for the second person. They waited for $t_{2}=10 \mathrm{min}$. Determine how much time $t_{1}$ passed from the start of the journey until the meeting. (15 points)
|
Answer: 20 min
Solution. Let $S$ be the distance between points A and B. We get $S=v_{1} t_{1}+v_{2} t_{1}$ (5 points). In addition, the same distance can be expressed as follows: $S=v_{2} t_{1}+v_{2} t_{1}+v_{2} t_{2}$ (5 points). As a result, we get $t_{1}=\frac{v_{2} t_{2}}{v_{1}-v_{2}}=\frac{4 \cdot 10}{6-4}=20 \text{min}$ (5 points).
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. A snail is crawling from one tree to another. In half a day, it crawled $l_{1}=4$ m. It realized it was all too much and turned back. It crawled $l_{2}=3$ m. It got tired. It fell asleep. The next day, everything repeated. And so every day. The distance between the trees is $s=40$ m. On which day of its journey will the snail reach the tree? (10 points)
|
Answer: on the 37th day
Solution. In one day, the snail advances towards the other tree by:
$\Delta l=l_{1}-l_{2}=1$ m (3 points). At a distance of $l_{1}=4$ m (i.e., one transition) from its goal, it will be after 36 days of travel (4 points). Therefore, the goal will be reached on the 37th day of travel (3 points).
## Tasks, answers, and evaluation criteria
6th grade
|
37
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A palindrome is a number that reads the same from left to right and from right to left. For example, the numbers 333 and $4884-$ are palindromes. It is known that a three-digit number $x$ is a palindrome. To it, 22 was added and a four-digit number was obtained, which is also a palindrome. Find $x$.
|
# Answer: 979.
Solution. Since $x+22$ is a four-digit number, the inequality $x+22 \geqslant 1000$ holds, from which $x \geqslant 978$. From the fact that $x$ is a three-digit palindrome, we get that this number starts and ends with the digit 9, and its middle digit is not less than 7. By checking the four possible options, we find the only answer.
Evaluation. If the answer is found but its uniqueness is not proven, 6 points. For a complete solution, 12 points.
|
979
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the cells of a $3 \times 3$ square, the numbers $0,1,2, \ldots, 8$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is 18?
|
Answer: 2.
Solution. Let's color the cells in a checkerboard pattern: let the corner and central cells be black, and the rest white. From the condition, it follows that cells of different colors contain numbers of different parity. Since there are five black cells and four white cells, we get that the even numbers are in the black cells. Their total sum is \(0+2+4+6+8=20\). Therefore, the number in the central cell is \(2=20-18\).
Evaluation. 13 points for a correct solution. If an example of the arrangement of numbers satisfying the problem's condition is provided but the uniqueness of the answer is not proven, 6 points.
## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences
## Final Stage
2017-2018 academic year
## Problems, Answers, and Evaluation Criteria
## 6th Grade Variant 2
physics
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. An aquarium in the shape of a rectangular parallelepiped has dimensions: length $2 m$, width - 600 mm, height - 60 cm. It is filled with water at a rate of 3 liters/minute. How many seconds after the start of filling will the aquarium be completely filled? (10 points)
|
Answer: 14400 s
Solution. Volume of the aquarium: $V=2 \cdot 0.6 \cdot 0.6=0.72 m^{3}$ (3 points). Filling rate: $v=\frac{0.003 m^{3}}{60 s}$ (3 points). The aquarium will be completely filled in: $t=\frac{V}{v}=\frac{0.72 \cdot 60}{0.003}=14400 s$ (4 points).
|
14400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Two people are walking towards each other from points $A$ and $B$ with speeds $v_{1}=6 \mathrm{~m} / \mathrm{c}$ and $v_{2}=4 \mathrm{~m} / \mathrm{c}$. At the moment of their meeting, one of them turned around and walked in the opposite direction, while the other continued in the same direction. The person who did not change direction arrived at the final point of their journey $t_{2}=10$ minutes earlier than the person who turned around. Determine how much time $t_{1}$ passed from the start of the journey until the meeting. (15 points)
|
Answer: 30 min
Solution. The distance from point $B$ to the meeting place $S=v_{2} t_{1}$ (5 points). In addition, the same distance can be expressed as follows: $S=v_{1}\left(t_{1}-t_{2}\right)$ (5 points). As a result, we get: $t_{1}=\frac{v_{1} t_{2}}{v_{1}-v_{2}}=\frac{6 \cdot 10}{6-4}=30$ min (5 points).
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. A snail is crawling from one tree to another. In half a day, it crawled $l_{1}=5$ m. It realized it was all too much and turned back. It crawled $l_{2}=4$ m. It got tired. It fell asleep. The next day, everything repeated. And so every day. The distance between the trees is $s=30$ m. On which day of its journey will the snail reach the tree? (10 points)
|
Answer: on the 26th day
Solution. In one day, the snail advances towards the other tree by:
$\Delta l=l_{1}-l_{2}=1$ m (3 points). At a distance of $l_{1}=5$ m (i.e., one transition) from its goal, it will be after 25 days of travel (4 points). Therefore, the goal will be reached on the 26th day of travel. (3 points).
|
26
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Let $a, b, c, d, e$ be positive integers. Their sum is 2018. Let $M=\max (a+b, b+c, c+d, d+e)$. Find the smallest possible value of $M$.
|
Answer: 673.
Solution. Estimation. We have the inequalities
$$
a+b \leqslant M ; \quad b+c \leqslant M ; \quad c+d \leqslant M ; \quad d+e \leqslant M
$$
Multiply the first and last inequalities by 2 and add the other two. We get:
$$
2(a+b+c+d+e)+b+d \leqslant 6 M ; \quad 6 M \geqslant 4036+b+d \geqslant 4038 ; \quad M \geqslant 673 .
$$
Example. The equality $M=673$ is achieved when $a=c=e=672$, $b=d=1$.
Evaluation. 14 points for a correct solution. If there is only a correct example, 7 points.
## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences Final Stage 2017-2018 academic year Problems, answers, and evaluation criteria 9th grade Variant 1 physics
|
673
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Thirteen identical metal rods are connected as follows (see fig.). It is known that the resistance of one rod \( R_{0}=10 \) Ohms. Determine the resistance of the entire structure if it is connected to a current source at points \( A \) and \( B \). (10 points)
|
Answer: $4 O m$
Solution. The equivalent circuit looks as follows (5 points),
where each of the resistors has a resistance of $R_{0}=10$ Ohms. As a result, the total resistance is: $R=\frac{4}{10} R_{0}=4$ Ohms (5 points).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. The specific heat capacity of a body with a mass of $m=2$ kg depends on the temperature as follows: $c=c_{0}(1+\alpha t)$, where $c_{0}=150$ J/kg ${ }^{\circ} \mathrm{C}-$ specific heat capacity at $0^{\circ} \mathrm{C}, \alpha=0.05^{\circ} \mathrm{C}^{-1}$ - temperature coefficient, $t$ - temperature in degrees Celsius. Determine the amount of heat that must be transferred to this body to heat it from $20^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. (10 points)
|
Answer: 96 kJ
Solution. Given that the specific heat capacity depends linearly on temperature, we can calculate its average value: $c_{c p}=\frac{c_{0}\left(1+\alpha t_{h}\right)+c_{0}\left(1+\alpha_{\kappa}\right)}{2}=600 \frac{\text { J }}{\text { kg } \cdot{ }^{\circ} \mathrm{C}} \quad(5$ points). The required amount of heat: $Q=c_{\varphi} m \Delta t=600 \cdot 2 \cdot 80=96000$ J = 96 kJ (5 points $)$.
|
96
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. A parallel beam of light falls normally on a thin lens. Behind the lens, at a distance of $80 \mathrm{~cm}$ from it, there is a screen on which a circular spot of a certain diameter is visible. If the screen is moved 40 cm, a spot of the same diameter will again be visible on the screen. Determine the focal length of the lens.
## (15 points)
|
Answer: 100 cm or 60 cm
Solution. A diagram explaining the situation described in the problem (5 points):
From this, it is clear that there are two possible scenarios: the screen can be moved away from the lens or moved closer to it. As a result, the focal length of the lens $F_{1}=80+20=100$ cm (5 points). Or $F_{2}=80-20=60$ cm (5 points).
## Final Stage
2017-2018 academic year
## Tasks, answers, and evaluation criteria
9th grade
Variant II
hatf Ћвtgr
r
|
100
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Let $a, b, c, d, e$ be positive integers. Their sum is 2345. Let $M=\max (a+b, b+c, c+d, d+e)$. Find the smallest possible value of $M$.
|
Answer: 782.
Solution. Estimation. We have the inequalities
$$
a+b \leqslant M ; \quad b+c \leqslant M ; \quad c+d \leqslant M ; \quad d+e \leqslant M
$$
Multiply the first and last inequalities by 2 and add the other two. We get:
$2(a+b+c+d+e)+b+d \leqslant 6 M ; \quad 6 M \geqslant 4690+b+d \geqslant 4692 ; \quad M \geqslant 782$.
Example. The equality $M=782$ is achieved when $a=c=e=781$, $b=d=1$.
Evaluation. 14 points for a correct solution. If there is only a correct example, 7 points.
## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences Final Stage 2017-2018 academic year Problems, answers, and evaluation criteria 9th grade Variant 2 physics
|
782
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. A parallel beam of light falls normally on a thin lens with a focal length of $F=150$ cm. Behind the lens, there is a screen on which a circular spot of a certain diameter is visible. If the screen is moved 40 cm, a spot of the same diameter will again be visible on the screen. Determine the initial distance from the lens to the screen. (15 points)
|
Answer: 170 cm or 130 cm
Solution. A diagram explaining the situation described in the problem (5 points):
From this, we can see that there are two possible scenarios: the screen can be moved away from the lens or moved closer to it. As a result, the initial distance from the lens to the screen:
$S_{1}=150+20=170 \mathrm{~cm} \quad\left(5\right.$ points) or $S_{2}=150-20=130 \mathrm{~cm}$ (5 points).
|
170
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the cells of a $3 \times 3$ square, the numbers $1,2,3, \ldots, 9$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$
|
Answer: 7.
Solution. Let's color the cells in a checkerboard pattern: suppose the corner and central cells are black, and the rest are white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white cells, we get that the odd numbers are in the black cells. Their total sum is $1+3+5+7+9=$ 25. Therefore, the number in the central cell is $7=25-18$.
Evaluation. 12 points for a correct solution. If an example of the arrangement of numbers satisfying the problem's condition is provided but the uniqueness of the answer is not proven, 6 points.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On a circle, 40 red points and one blue point are marked. All possible polygons with vertices at the marked points are considered. Which type of polygons is more numerous, and by how many: those with a blue vertex, or those without it?
|
Answer: The number of polygons with a blue vertex is 780 more than the number of polygons without a blue vertex.
Solution. Let's call polygons with a blue vertex blue, and those without a blue vertex red. Take an arbitrary red polygon. Adding a blue vertex to it gives exactly one blue polygon. Thus, any blue polygon with at least four vertices can be obtained this way. Therefore, the difference between the number of blue polygons and the number of red polygons is equal to the number of blue triangles. The number of blue triangles is the same as the number of ways to choose two red points (two red vertices of a blue triangle), i.e., $\frac{40 \cdot 39}{2} = 780$.
Evaluation. 14 points for a correct solution.
## Tasks, answers, and evaluation criteria
7th grade
Variant 1 physics
|
780
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A person is walking parallel to a railway track at a constant speed. A train also passes by him at a constant speed. The person noticed that depending on the direction of the train, it passes by him either in $t_{1}=1$ minute or in $t_{2}=2$ minutes. Determine how long it would take the person to walk from one end of the train to the other.
## $(15$ points)
|
Answer: 4 min
Solution. When the train and the person are moving towards each other:
$l=\left(v_{n}+v_{u}\right) \cdot t_{1}$ (3 points), where $l$ - the length of the train, $v_{n}$ - its speed, $v_{u}$ - the speed of the person. If the directions of the train and the person are the same, then:
$l=\left(v_{n}-v_{u}\right) \cdot t_{2} \quad\left(3\right.$ points). In the situation where the person is walking on the train: $l=v_{u} \cdot t_{3}$. (3 points). Solving this system of equations, we get the final answer:
$t_{3}=\frac{2 t_{1} t_{2}}{t_{2}-t_{1}}=\frac{2 \cdot 1 \cdot 2}{2-1}=4$ minutes (6 points).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. At the ends of a vertically positioned homogeneous spring, two small loads are fixed. Above is a load with mass $m_{1}$, and below is $-m_{2}$. A person grabbed the middle of the spring and held it vertically in the air. In this case, the upper half of the spring was deformed by $x_{1}=8 \mathrm{~cm}$, and the lower half by $x_{2}=15 \mathrm{~cm}$. After that, he placed the spring on a horizontal surface without flipping it and released it. Determine the magnitude of the spring's deformation in this case. Neglect the size of the person's hand compared to the length of the spring.
## $(15$ points)
|
Answer: 16 cm
Solution. If the stiffness of the spring is $k$, then the stiffness of half the spring is $2 k$ (4 points). For the situation where the spring is held by a person: $m_{1} g=2 k x_{1}$ (4 points).
For the situation where the spring stands on a surface: $m_{1} g=k x$ (4 points). We obtain that $x=2 x_{1}=16$ cm (3 points).
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the cells of a $3 \times 3$ square, the numbers $0,1,2, \ldots, 8$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$
|
Answer: 2.
Solution. Let's color the cells in a checkerboard pattern: suppose the corner and central cells are black, and the rest are white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white cells, we get that the even numbers are in the black cells. Their total sum is $0+2+4+6+8=20$. Therefore, the number in the central cell is $2=20-18$.
Evaluation. 12 points for a correct solution. If an example of the arrangement of numbers satisfying the problem's condition is provided but the uniqueness of the answer is not proven, 6 points.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On a circle, 60 red points and one blue point are marked. All possible polygons with vertices at the marked points are considered. Which type of polygons is more numerous, and by how many: those with a blue vertex, or those without it?
|
Answer: The number of polygons with a blue vertex is 1770 more than the number of polygons without a blue vertex.
Solution. Let's call polygons with a blue vertex blue, and those without a blue vertex red. Take an arbitrary red polygon. Adding a blue vertex to it gives exactly one blue polygon. Thus, any blue polygon with at least four vertices can be obtained this way. Therefore, the difference between the number of blue polygons and the number of red polygons is equal to the number of blue triangles. The number of blue triangles is the same as the number of ways to choose two red points (two red vertices of a blue triangle), i.e., $\frac{60 \cdot 59}{2}=1770$.
Evaluation. 14 points for a correct solution.
## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences
## Final Stage 2017-2018 academic year
## Problems, Answers, and Evaluation Criteria
## 7th Grade Variant 2
physics5. A person is walking parallel to a railway track at a constant speed. A train also passes by at a constant speed. The person noticed that depending on the direction of the train's movement, it passes by in either $t_{1}=2 \text{ min}$ or $t_{2}=4$ min. Determine how long it would take the person to walk from one end of the train to the other.
## (15 points)
Answer: 8 min
Solution. When the train and the person are moving towards each other, $l=\left(v_{n}+v_{4}\right) \cdot t_{1} \quad$ (3 points), where $l$ is the length of the train, $v_{n}$ is its speed, and $v_{4}$ is the person's speed. If the directions of the train and the person are the same, then $l=\left(v_{n}-v_{4}\right) \cdot t_{2} \quad$ (3 points). In the situation where the person is walking along the train: $l=v_{u} \cdot t_{3}$ (3 points). Solving this system of equations, we get the final answer: $t_{3}=\frac{2 t_{1} t_{2}}{t_{2}-t_{1}}=\frac{2 \cdot 2 \cdot 4}{4-2}=8$ minutes (6 points).
|
1770
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The cold water tap fills the bathtub in 17 minutes, and the hot water tap in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water in it?
|
Answer: in 3 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 8.5 minutes. Therefore, the hot water tap should be open for 3 minutes longer.
Evaluation. 12 points for the correct solution.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the sum of all four-digit numbers in which the digits $0,3,6,9$ do not appear.
|
Answer: $9999 \cdot 6^{4} / 2=6479352$.
Solution. All four-digit numbers that meet the condition of the problem are divided into pairs of numbers of the form (1111,8888), (1112,8887), (1113,8886), ..., (4555,5444). In each pair, the sum of the numbers is 9999. Let's count the number of pairs. In total, there are $6^{4}$ four-digit numbers of the specified type (since each digit can be chosen in 4 ways). Therefore, there are half as many pairs. The sum of all numbers is $9999 \cdot 6^{4} / 2=6479352$.
Grading. 12 points for a correct solution. The answer can also be presented as the formula $9999 \cdot 6^{4} / 2$. If the formula is correct but there are arithmetic errors in the calculation using this formula, 10 points.
|
6479352
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. What is the greatest length that a closed, non-self-intersecting broken line can have, which runs along the grid lines of a square grid of size $8 \times 8$?
#
|
# Answer: 80.
Solution. We will color the nodes of the grid in a checkerboard pattern, with black and white colors. The length of a closed non-self-intersecting broken line is equal to the number of nodes it passes through. Each segment of the broken line connects a black and a white node. When traversing the broken line, the colors of the nodes alternate, so the length of the closed broken line is an even number. Since there are 81 nodes in total in the grid, the length of the broken line is at most 80. A corresponding example is easily constructed (Fig. 1).
Fig. 1
Remark. Another reasoning is possible. Let's start constructing a closed broken line starting from some node. Since we need to return to it, the number of horizontal steps to the right is the same as the number of steps to the left. Therefore, the number of horizontal steps is even. The same is true for vertical steps. Hence, we again obtain that the length of the closed broken line (whose segments follow the grid lines) is an even number.
Evaluation. 14 points for a correct solution. 1 point if only the answer is given. 4 points if an example of a broken line of length 80 is provided, but there is no estimation of the length of the broken line.
2016-2017 academic year
\section*{Tasks, answers, and evaluation criteria for 7th grade
|
80
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The cold water tap fills the bathtub in 19 minutes, and the hot water tap in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water?
|
Answer: in 2 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 9.5 minutes. Therefore, the hot water tap should be open for 2 minutes longer.
Evaluation. 12 points for the correct solution.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the sum of all four-digit numbers in which the digits $0,4,5,9$ do not appear.
|
Answer: $9999 \cdot 6^{4} / 2=6479352$.
Solution. All four-digit numbers that meet the condition of the problem are divided into pairs of numbers of the form (1111,8888$),(1112,8887),(1113,8886)$, $\ldots$... (4555,5444$)$. In each pair, the sum of the numbers is 9999. Let's count the number of pairs. In total, there are $6^{4}$ four-digit numbers of the specified type (since each digit can be chosen in 4 ways). Therefore, there are half as many pairs. The sum of all numbers is $9999 \cdot 6^{4} / 2=6479352$.
Grading. 12 points for a correct solution. The answer can also be presented as the formula $9999 \cdot 6^{4} / 2$. If the formula is correct but there are arithmetic errors in the calculation using this formula, 10 points.
|
6479352
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. What is the greatest length that a closed, non-self-intersecting broken line can have, which runs along the grid lines of a $6 \times 10$ cell field?
|
Answer: 76.
Solution. We will color the nodes of the grid in a checkerboard pattern, with black and white colors. The length of a closed non-self-intersecting broken line is equal to the number of nodes it passes through. Each segment of the broken line connects a black and a white node. When traversing the broken line, the colors of the nodes alternate, so the length of the closed broken line is an even number. Since there are 77 nodes in total in the grid, the length of the broken line is at most 76. A corresponding example is easily constructed (Fig. 2).
Fig. 2
Remark. Another reasoning is possible. Let's start constructing a closed broken line from some node. Since we need to return to it, the number of horizontal steps to the right is the same as the number of steps to the left. Therefore, the number of horizontal steps is even. The same is true for vertical steps. Hence, we again obtain that the length of the closed broken line (whose segments follow the grid lines) is an even number.
Evaluation. 14 points for a correct solution. 1 point if only the answer is given. 4 points if an example of a broken line of length 76 is provided, but there is no estimation of the length of the broken line.
|
76
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given a parallelogram $A B C D$. It is known that the centers of the circles circumscribed around triangles $A B C$ and $C D A$ lie on the diagonal $B D$. Find the angle $D B C$, if $\angle A B D=40^{\circ}$.
|
Answer: $50^{\circ}$ or $40^{\circ}$.
Solution. The center of the circle circumscribed around triangle $A B C$ lies on the perpendicular bisector of segment $A C$. Therefore, either this center is the midpoint of $A C$ (and then $A B C D$ is a rectangle), or $D B \perp A C$ (and then $A B C D$ is a rhombus). In the first case, angle $D B C$ complements angle $A B D$ to a right angle, and in the second case, these angles are equal to each other.
Evaluation. 10 points for a correct solution. If only the case of the rhombus is considered, 6 points. If only the case of the rectangle is considered, 4 points.
|
50
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given a parallelogram $A B C D$. It is known that the centers of the circles circumscribed around triangles $A B C$ and $C D A$ lie on the diagonal $B D$. Find the angle $D B C$, if $\angle A B D=35^{\circ}$.
|
Answer: $55^{\circ}$ or $35^{\circ}$.
Solution. The center of the circle circumscribed around triangle $A B C$ lies on the perpendicular bisector of segment $A C$. Therefore, either this center is the midpoint of $A C$ (and then $A B C D$ is a rectangle), or $D B \perp A C$ (and then $A B C D$ is a rhombus). In the first case, angle $D B C$ complements angle $A B D$ to a right angle, and in the second case, these angles are equal to each other.
Evaluation. 10 points for a correct solution. If only the case of the rhombus is considered, 6 points. If only the case of the rectangle is considered, 4 points.
|
55
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Solve the equation
$$
2 x+2+x \sqrt{x^{2}+1}+(x+2) \sqrt{x^{2}+4 x+5}=0 .
$$
|
Answer: -1.
Solution. Let $f(x)=x\left(1+\sqrt{x^{2}+1}\right)$. The original equation can be rewritten as $f(x)+f(x+2)=0$. Note that the function $f(x)$ is odd. It is increasing on the positive half-axis (as the product of positive increasing functions). Due to its oddness, it is increasing on the entire real line. Further, we have
$$
f(x)=-f(x+2) ; \quad f(x)=f(-x-2)
$$
Since an increasing function takes each of its values exactly once, it follows that $x=-x-2$, from which $x=-1$.
Remark. The monotonicity of the left side of the original equation could also have been established using the derivative.
Evaluation. 13 points for a correct solution. If the answer is guessed but not proven that there are no other solutions, 3 points.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (17 points) Three cyclists, Dima, Misha, and Petya, started a 1 km race simultaneously. At the moment Misha finished, Dima still had to cover one tenth of the distance, and at the moment Dima finished, Petya still had to cover one tenth of the distance. How far apart (in meters) were Petya and Dima when Misha finished? (The cyclists are riding at constant speeds).
|
Answer: 90.
Solution. Let $v_{1}$ be the speed of Misha, $v_{2}$ be the speed of Dima, and $v_{3}$ be the speed of Petya. Then $v_{2}=0.9 v_{1}$, since in the same amount of time Misha traveled 1000 m, while Dima traveled 900 m, and $v_{3}=0.9 v_{2}$, since in the same amount of time Dima traveled 1000 m, while Petya traveled 900 m. Since $v_{3}=$ $0.9 \cdot 0.9 v_{1}=0.81 v_{1}$, at the moment Misha finished, Petya had traveled 810 meters. The distance between Petya and Dima: $900-810=90$.
|
90
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) Point $M$ lies inside segment $A B$, the length of which is 40 cm. Points are chosen: $N$ at the midpoint of $A M, P$ at the midpoint of $M B, C$ at the midpoint of $N M, D$ at the midpoint of $M P$. Find the length of segment $C D$ in cm.
|
Answer: 10.
## Solution.
$$
\begin{aligned}
& N P=N M+M P=\frac{1}{2} A M+\frac{1}{2} M B=\frac{1}{2} A B=20, \\
& C D=C M+M D=\frac{1}{2} N M+\frac{1}{2} M P=\frac{1}{2} N P=10 .
\end{aligned}
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (16 points) In a class, some students study only English, some study only German, and some study both languages. What percentage of the class studies both languages if 90% of all students study English and 80% study German?
|
Answer: 70.
Solution. From the condition, it follows that $10 \%$ do not study English and $20 \%$ do not study German. Therefore, $10 \%+20 \%=30 \%$ study only one language, and the remaining $70 \%$ study both languages.
|
70
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (17 points) Three cyclists, Dima, Misha, and Petya, started a 1 km race simultaneously. At the moment Misha finished, Dima still had to cover one tenth of the distance, and at the moment Dima finished, Petya still had to cover one tenth of the distance. How far apart (in meters) were Petya and Misha when Misha finished? (The cyclists are riding at constant speeds).
|
Answer: 190.
Solution. Let $v_{1}$ be the speed of Misha, $v_{2}$ be the speed of Dima, and $v_{3}$ be the speed of Petya. Then $v_{2}=0.9 v_{1}$, since in the same time Misha traveled $1000 \mathrm{m}$, while Dima traveled $900 \mathrm{m}$, and $v_{3}=0.9 v_{2}$, since in the same time Dima traveled 1000 m, while Petya traveled 900 m. Since $v_{3}=$ $0.9 \cdot 0.9 v_{1}=0.81 v_{1}$, at the moment Misha finished, Petya had traveled 810 meters. The distance between Petya and Misha:
$1000-810=190$
|
190
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) Point $M$ lies inside segment $A B$, the length of which is 60 cm. Points are chosen: $N$ at the midpoint of $A M, P$ at the midpoint of $M B, C$ at the midpoint of $N M, D$ at the midpoint of $M P$. Find the length of segment $C D$.
|
Answer: 15 cm.
## Solution.
$$
\begin{aligned}
& \quad \dot{N} \quad \dot{C} \quad \dot{M} \dot{D} \cdot B \\
& N P=N M+M P=\frac{1}{2} A M+\frac{1}{2} M B=\frac{1}{2} A B=30 \\
& C D=C M+M D=\frac{1}{2} N M+\frac{1}{2} M P=\frac{1}{2} N P=15
\end{aligned}
$$
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (16 points) In a class, some students study only English, some study only German, and some study both languages. What percentage of the class studies both languages if 80% of all students study English and 70% study German?
|
Answer: 50.
Solution. From the condition, it follows that $20 \%$ do not study English and $30 \%$ do not study German. Therefore, $20 \%+30 \%=50 \%$ study only one language, and the remaining $50 \%$ study both languages.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem № 6 (10 points)
Five identical balls are rolling towards each other on a smooth horizontal surface. The speeds of the first and second are \( v_{1} = v_{2} = 0.5 \) m/s, while the others are \( v_{3} = v_{4} = v_{5} = 0.1 \) m/s. The initial distances between the balls are the same, \( l = 2 \) m. All collisions are perfectly elastic. How much time will pass between the first and last collisions in this system?
1
2
3
4
5
## Time: 10 min
#
|
# Solution and Evaluation Criteria:
In the case of a perfectly elastic collision, identical balls "exchange" velocities.
Therefore, the situation can be considered as if the balls pass through each other with unchanged speeds. The first collision occurs between the second and third balls. The last collision will occur at the moment when the first ball "passes" by the fifth.
(4 points)
The corresponding time: \( t = \frac{3 l}{v_{1} + v_{5}} = \frac{3 \cdot 2}{0.6} = 10 \text{s} \)
#
|
10
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (17 points) Solve the equation ||$|x-1|+2|-3|=-2 x-4$.
|
Answer: -4.
Solution. The set of possible solutions for the inequality $-2 x-4 \geq 0$, that is, $x \leq-2$. Under this condition, the inner absolute value is uniquely determined. Using the property of the absolute value $|-a|=|a|$, we get the equation $||x-3|-3|=-2 x-4$. Similarly, we expand the inner absolute value, obtaining the equation $|x|=-2 x-4$. Under the obtained restriction, the last absolute value is uniquely determined, resulting in the equation $-x=-2 x-4$, that is, $x=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (16 points) Aunt Masha decided to bake a cake. She kneaded the dough, which according to the recipe contains flour, butter, and sugar in the weight ratio of $3: 2: 1$, and mixed butter with sugar for the cream in the ratio of 2:3. Changing her mind about baking the cake, she combined both mixtures, added 200 g of flour, and prepared dough for cookies, in which flour, butter, and sugar are in the ratio of 5:3:2. How much butter did Aunt Masha use? Write the answer in grams.
|
Answer: 480.
Solution. Let the cake dough contain flour, butter, and sugar in the amounts of $3 x$, $2 x$, and $x$ respectively, and the cream contain butter and sugar in the amounts of $2 y$ and $3 y$ respectively. Using the ratios for the cookies, we get the system of equations $\left\{\begin{array}{l}\frac{3 x+200}{2 x+2 y}=\frac{5}{3}, \\ \frac{2 x+2 y}{x+3 y}=\frac{3}{2} .\end{array}\right.$ After transformations, we get the system $\left\{\begin{array}{c}x+10 y=600, \\ x-5 y=0 .\end{array}\right.$ From which we find that $x=200, y=40$. Therefore, Aunt Masha spent $2 x+2 y=400+80=480$ grams of butter.
|
480
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (15 points) A ball was thrown vertically upwards from a balcony. It is known that it hit the ground after 6 seconds. Given that the initial speed of the ball is 20 m/s, determine the height of the balcony. The acceleration due to gravity is 10 m/s².
|
Answer: 60 m.
Solution. The equation of motion of the ball:
$$
y=0=h_{0}+v_{0} t-\frac{g t^{2}}{2}=h_{0}+20 \cdot 6-\frac{10 \cdot 6^{2}}{2}
$$
We get $h_{0}=60 m$ m.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (20 points) Two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface.
The refractive index of the glass $n=1.5$. The angle of refraction for the first ray $\beta=25^{\circ}$. Determine the angle between the refracted rays.
|
Answer: $56^{\circ}$.
Solution. The law of refraction for the first ray $\frac{\sin \alpha}{\sin 25^{\circ}}=1.5$. Therefore, the angle of incidence of the first ray $\alpha=39.34^{\circ}$. The angle of incidence of the second ray $\beta=90^{\circ}-39.34^{\circ}=$ $50.66^{\circ}$. The law of refraction for the second ray $\frac{\sin 50.66^{\circ}}{\sin \gamma}=1.5$. The angle of refraction of the second ray $\gamma=31^{\circ}$. The angle between the refracted rays: $\beta+\gamma=56^{\circ}$.
|
56
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (17 points) Solve the equation ||$|x-2|+3|-4|=-3 x-9$.
|
Answer: -5.
Solution. The set of possible solutions for the inequality $-3 x-9 \geq 0$, that is, $x \leq-3$. Under this condition, the inner absolute value is uniquely determined. Using the property of absolute value $|-a|=|a|$, we get the equation $||x-5|-4|=-3 x-9$. Similarly, we expand the inner absolute value, obtaining the equation $|x-1|=-3 x-9$. Under the obtained restriction, the last absolute value is uniquely determined, resulting in the equation $-x+1=-3 x-9$, that is, $x=-5$.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (16 points) Aunt Masha decided to bake a cake. She mixed the dough, which according to the recipe includes flour, butter, and sugar in the weight ratio of $3: 2: 1$, and mixed butter with sugar for the cream in the ratio of $2: 3$. Changing her mind about baking the cake, she combined both mixtures, added 300 g of flour, and prepared dough for cookies, in which flour, butter, and sugar are in the ratio of 5:3:2. How much flour did Aunt Masha use? Write the answer in grams.
|
Answer: 1200.
Solution. Let the dough for the cake contain flour, butter, and sugar in the amounts of $3x$, $2x$, and $x$ respectively, and the cream contain butter and sugar in the amounts of $2y$ and $3y$ respectively. Using the ratios for the cookies, we get the system of equations $\left\{\begin{array}{l}\frac{3x+300}{2x+2y}=\frac{5}{3}, \\ \frac{2x+2y}{x+3y}=\frac{3}{2} .\end{array}\right.$
We find that $x=300, y=60$. Thus, Aunt Masha spent $3x+300=900+300=1200$ grams of flour.
3. (17 points) In a right triangle, the legs are 5 and 12. Find the distance between the centers of the inscribed and circumscribed circles. Write the square of this distance in the answer.
Answer: 16.25.
Solution. By the Pythagorean theorem, the hypotenuse is 13, and the area of the triangle is $S=\frac{1}{2} \cdot 5 \cdot 12=30$. The semiperimeter $p=\frac{5+12+13}{2}=15$, so the radius of the inscribed circle $r=\frac{s}{p}=2$. Let $O_{1}$ be the center of the inscribed circle, $O_{1}P$ and $O_{1}K$ be the radii of the inscribed circle, drawn to the leg and the hypotenuse at the points of tangency of the circle, with legs $BC=12, AC=5$ by the problem statement. We have $BP=BC-r=12-2=10$. By the property of tangents drawn from one point to a circle, $BK=BP=10$. The center of the circumscribed circle $O_{2}$ around the right triangle lies at the midpoint of the hypotenuse and $BO_{2}=6.5$.
In the right triangle $KO_{1}O_{2}$, we have $KO_{2}=10-6.5=3.5, KO_{1}=r=2, O_{1}O_{2}$ is the desired distance. By the Pythagorean theorem, $O_{1}O_{2}^2=4+12.25=16.25$.
|
1200
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (20 points) Two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface. The refractive index of the glass $n=1.6$. The angle of refraction for the first ray $\beta=30^{\circ}$. Determine the angle between the refracted rays.
|
Answer: $52^{\circ}$.
Solution. The law of refraction for the first ray $\frac{\sin \alpha}{\sin 30^{\circ}}=1.6$. Therefore, the angle of incidence of the first ray $\alpha=53.13^{0}$. The angle of incidence of the second ray $\beta=90^{0}-53.13^{0}=$ $36.87^{\circ}$. The law of refraction for the second ray $\frac{\sin 36.87^{\circ}}{\sin \gamma}=1.6$. The angle of refraction of the second ray $\gamma=22^{0}$. In the end, we get the angle between the refracted rays $\beta+$ $\gamma=52^{0}$.
|
52
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (15 points) Two springs with stiffnesses of $6 \mathrm{kH} / \mathrm{m}$ and $12 \mathrm{kH} / \mathrm{m}$ are connected in series. How much work is required to stretch this system by 10 cm.
|
Answer: 20 J.
Solution. Total stiffness: $k=\frac{k_{1} k_{2}}{k_{1}+k_{2}}=\frac{6 \cdot 12}{6+12}=4 \mathrm{kH} / \mathrm{m}$. Work:
$$
A=\Delta W=\frac{k x^{2}}{2}=\frac{4000 \cdot \cdot 0.1^{2}}{2}=20 \text { J. }
$$
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (12 points) Solve the equation
$$
\sqrt[3]{(7-x)^{2}}-\sqrt[3]{(7-x)(9+x)}+\sqrt[3]{(9+x)^{2}}=4
$$
|
Answer: -1
Solution. Let $a=7-x, b=9+x$. Notice that $a+b=16$. We obtain the system of equations
$$
\left\{\begin{array}{c}
\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}}=4 \\
a+b=16
\end{array}\right.
$$
Multiply the first equation by $\sqrt[3]{a}+\sqrt[3]{b}$, using the formula for the difference of cubes, we get $\sqrt[3]{a}+\sqrt[3]{b}=4$. Square this equality, and subtract the first equation of the system from the result, we get $\sqrt[3]{a b}=4$.
The system $\left\{\begin{array}{c}\sqrt[3]{a b}=4, \\ a+b=16,\end{array}\right.$ reduces to a quadratic equation. Solving it, we get $a=8 ; x=-1$. Let's check the solution.
Evaluation criteria. Full justified solution - 12 points. Correct answer obtained without justification (guessed) - 3 points.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (12 points) During the break, a fly flew into the math classroom and started crawling on a poster, on which the graph of a quadratic function $y=f(x)$ was depicted in the coordinate plane, with the leading coefficient equal to 1. Initially, the fly moved exactly along the parabola up to the point with an abscissa equal to 2, but then it started moving along a straight line until it hit the parabola again at the point with an abscissa equal to 4. Find $f(3)$, given that the line $y=2023 x$ intersects the fly's path along the straight line segment at its midpoint.
|
Answer: 6068.
Solution. Let the quadratic function be of the form $y=x^{2}+b x+c$. The midpoint of the line segment has coordinates $\left(\frac{2+4}{2} ; \frac{f(2)+f(4)}{2}\right)$, on the other hand, $\left(3 ; 6069\right)$. Since $f(2)=4+2 b+c, f(4)=16+4 b+c$, then $20+6 b+2 c=12138$ or $3 b+c=6059$. Therefore, $f(3)=9+3 b+c=9+6059=6068$.
Grading criteria. Full points for a complete and justified solution - 12 points. For a correct approach with arithmetic errors, minus 3 points.
|
6068
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (13 points) In a dance ensemble, there are 8 boys and 16 girls. Some of them form mixed (boy and girl) dance pairs. It is known that in each pair, at least one of the partners does not belong to any other pair. What is the maximum number of dance pairs that can be formed in this ensemble?
|
Answer: 22.
Solution. Example. Let Yana and Maxim be the best dancers in the ensemble. If Maxim dances with all the girls except Yana, and Yana dances with all the boys except Maxim, then the condition of the problem is satisfied and the number of pairs will be $15+7=22$.
Estimate. Let's call a pair in which the partner does not participate in other pairs - male, and a pair in which the partner does not participate in other pairs - female. (A pair can be both male and female). If there are 8 male pairs, then there are no other pairs. Therefore, if we want the total number of pairs to be more than 8, the number of male pairs should not exceed 7. If there are 16 female pairs, then there are also no other pairs. And if we want the total number of pairs to be more than 16, the number of female pairs should not exceed 15. Thus, the total number of pairs (each of which is either male or female) is no more than $7+15=22$.
Grading criteria. Full solution - 13 points. Only the example is worth 6 points; only the estimate - 7 points.
|
22
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (12 points) Solve the equation
$$
\sqrt[3]{(9-x)^{2}}-\sqrt[3]{(9-x)(7+x)}+\sqrt[3]{(7+x)^{2}}=4
$$
|
Answer: 1.
Solution. Let $a=9-x, b=7+x$. Note that $a+b=16$. We obtain the system of equations
$$
\left\{\begin{array}{c}
\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}}=4 \\
a+b=16
\end{array}\right.
$$
Multiply the first equation by $\sqrt[3]{a}+\sqrt[3]{b}$, using the formula for the difference of cubes, we get $\sqrt[3]{a}+\sqrt[3]{b}=4$. Square this equality, and subtract the first equation of the system from the result, we get $\sqrt[3]{a b}=4$.
The system $\left\{\begin{array}{c}\sqrt[3]{a b}=4, \\ a+b=16\end{array}\right.$ reduces to a quadratic equation. Solving it, we get $a=8 ; x=1$. Let's check the solution.
Evaluation criteria. Full justified solution - 12 points. Correct answer obtained without justification (guessed) - 3 points.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (12 points) During the break, a fly flew into the math classroom and started crawling on a poster, on which the graph of a quadratic function $y=f(x)$ was depicted in the coordinate plane, with the leading coefficient equal to -1. Initially, the fly moved exactly along the parabola up to the point with an abscissa of 2, but then it started moving along a straight line until it hit the parabola again at the point with an abscissa of 4. Find $f(3)$, given that the line $y=2023 x$ intersects the fly's path along the straight line segment at its midpoint.
|
Answer: 6070.
Solution. Let the quadratic function be of the form $y=-x^{2}+b x+c$. The midpoint of the line segment has coordinates $\left(\frac{2+4}{2} ; \frac{f(2)+f(4)}{2}\right)$, and on the other hand, $(3 ; 6069)$. Since $f(2)=-4+2 b+c, f(4)=-16+4 b+c$, then $-20+6 b+2 c=12138$ or $3 b+c=6079$. Therefore, $f(3)=-9+3 b+c=-9+6079=6070$.
Grading criteria. Full points for a complete and justified solution - 12 points. For a correct approach with arithmetic errors, minus 3 points.
|
6070
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (13 points) In a dance ensemble, there are 8 boys and 20 girls. Some of them form mixed (boy and girl) dance pairs. It is known that in each pair, at least one of the partners does not belong to any other pair. What is the maximum number of dance pairs that can be formed in this ensemble?
|
Answer: 26.
Solution. Example. Let Yana and Maxim be the best dancers in the ensemble. If Maxim dances with all the girls except Yana, and Yana dances with all the boys except Maxim, then the condition of the problem is satisfied and the number of pairs will be $19+7=26$.
Estimate. Let's call a pair in which the partner does not participate in other pairs - male, and a pair in which the partner does not participate in other pairs - female. (A pair can be both male and female). If there are 8 male pairs, then there are no other pairs. Therefore, if we want the total number of pairs to be more than 8, the number of male pairs should not exceed 7. If there are 20 female pairs, then there are also no other pairs. And if we want the total number of pairs to be more than 20, the number of female pairs should not exceed 19. Thus, the total number of pairs (each of which is either male or female) is no more than $7+19=26$.
Grading criteria. Full solution - 13 points. Only the example is worth 6 points; only the estimate - 7 points.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (15 points) A mosquito was moving over the water in a straight line at a constant speed of \( v = 0.5 \) m/s and at the end of its movement, it landed on the water surface. 20 seconds before landing, it was at a height of \( h = 6 \) m above the water surface. The cosine of the angle of incidence of the sunlight on the water surface is 0.6. The incident sunlight, which creates the shadow of the mosquito, and its trajectory lie in the same vertical plane. Determine the speed at which the shadow of the mosquito moved along the bottom of the water body.
|
Answer: 0 m/s or $0.8 \mathrm{~m} / \mathrm{s}$.
Solution. The mosquito flew a distance of $s=v t=0.5 \cdot 20=10$ m before landing.
That is, it moved along the trajectory $A B$.
The cosine of the angle between the trajectory and the water surface is 0.8.
Obviously, the speed of the shadow moving along the bottom coincides with the speed of the shadow moving on the surface of the pond.
Since the angle of incidence of the sunlight coincides with the angle between the trajectory and the water surface, there are two possible scenarios.
The first - the mosquito is moving along the sunbeam.
The speed of the shadow on the bottom of the pond $v_{\text {shadow }}=0 \mathrm{~m} / \mathrm{s}$.
The second - the mosquito is flying towards the sunbeam. 2
The speed of the shadow $v_{\text {shadow }}=2 v \cos \beta=2 \cdot 0.5 \cdot 0.8=0.8 \mathrm{~m} / \mathrm{s}$.
|
0
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (13 points) What is the minimum number of participants that could have been in the school drama club if the number of fifth graders was more than $25 \%$ but less than $35 \%$, the number of sixth graders was more than $30 \%$ but less than $40 \%$, and the number of seventh graders was more than $35 \%$ but less than $45 \%$ (there were no participants from other grades).
|
Answer: 11. Fifth-graders -3, sixth-graders -4, seventh-graders - 4.
Solution. Let the required number of participants be
$$
n=a+b+c
$$
where $a$ is the number of fifth-graders, $b$ is the number of sixth-graders, $c$ is the number of seventh-graders. Note: $a<c$ and all three numbers constitute more than 0.25 and less than 0.5 of the total number of participants according to the condition. Consider all remaining cases of the sum $(*)$ taking into account these observations.
$5=1+2+2$. The condition for sixth-graders is not met.
$7=2+2+3$. The condition for sixth-graders is not met.
$8=2+3+3$. The condition for fifth-graders is not met.
$9=3+3+3$. The condition $a<c$ is not met.
$10=3+3+4$. The condition for sixth-graders is not met.
$11=3+4+4$. All conditions are met.
Evaluation criteria. Correct answer based on a complete enumeration without losing cases - 13 points. Correct answer, but not all possible cases are considered - 7 points. For the correct answer without explanation 2 points.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (15 points) The density of a body $\rho$ is defined as the ratio of the body's mass $m$ to its volume $V$. A unit of mass used in jewelry is the carat (1 carat equals 0.2 grams). A unit of length used in many countries is the inch (1 inch equals 2.54 centimeters). It is known that the density of diamond is $\rho=3.5 \mathrm{r} / \mathrm{c}^{3}$. Convert this value to carats per cubic inch.
|
Answer: $\approx 287$ carats/inch ${ }^{3}$.
Solution. 1 gram $=\frac{1}{0.2}$ carat $=5$ carats,
1 cm $=\frac{1}{2.54}$ inch.
We get: $\rho=3.5 \frac{\mathrm{r}}{\mathrm{cm}^{3}}=3.5 \frac{5 \text { carats }}{\left(\frac{1}{2.54} \text { inch }\right)^{3}}=3.5 \cdot 5 \cdot 2.54^{3} \frac{\text { carats }}{\text { inch }{ }^{3}} \approx 287 \frac{\text { carats }}{\text { inch }{ }^{3}}$.
(5 points)
|
287
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) An industrial robot travels from point $A$ to point $B$ according to a pre-determined algorithm. The diagram shows a part of its repeating trajectory. Determine how many times faster it would reach from point $A$ to point $B$ if it moved in a straight line at three times the speed?
|
Answer: 9 times.
Solution. The time spent by the robot moving along the given trajectory:
$t_{1}=N \frac{2 a+3 a+a+3 a}{v}=9 \frac{N a}{v}$.
The time spent by the robot moving in a straight line:
$t_{2}=N \frac{2 a+a}{3 v}=\frac{N a}{v}$.
We obtain that the robot will arrive $\frac{t_{1}}{t_{2}}=9$ times faster.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (13 points) What is the minimum number of participants that could have been in the school drama club if the number of fifth graders was more than $22 \%$ but less than $27 \%$, the number of sixth graders was more than $25 \%$ but less than $35 \%$, and the number of seventh graders was more than $35 \%$ but less than $45 \%$ (there were no participants from other grades).
|
Answer: 9. Fifth-graders - 2, sixth-graders - 3, seventh-graders - 4.
Solution. Let the required number of participants be
$$
n=a+b+c
$$
where $a$ is the number of fifth-graders, $b$ is the number of sixth-graders, and $c$ is the number of seventh-graders. Note: $a<c$ and all three numbers constitute more than 0.2 and less than 0.5 of the total number of participants according to the condition. Consider all remaining sum cases ( * with these notes in mind.
$5=1+2+2$. The condition for sixth-graders is not met.
$7=2+2+3$. The condition for fifth-graders is not met.
$8=2+3+3$. The condition for sixth-graders is not met.
$9=2+3+4$. All conditions are met.
Evaluation criteria. Correct answer based on a complete enumeration without losing cases - 13 points. Correct answer, but not all possible cases are considered - 7 points.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (15 points) The density of a body $\rho$ is defined as the ratio of the body's mass $m$ to its volume $V$. A unit of mass used in jewelry is the carat (1 carat equals 0.2 grams). A unit of length used in many countries is the inch (1 inch equals 2.54 centimeters). It is known that the density of emerald is $\rho=2.7 \text{ g} / \mathrm{cm}^{3}$. Convert this value to carats per cubic inch.
|
Answer: $\approx 221$ carats/inch ${ }^{3}$.
Solution. 1 gram $=\frac{1}{0.2}$ carat $=5$ carats,
1 cm $=\frac{1}{2.54}$ inch.
(2 points)
|
221
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (10 points) A body moves along the Ox axis. The dependence of velocity on time is shown in the figure. Determine the distance traveled by the body in 6 seconds.
|
Answer: 5 meters.
Solution. In the first second, the body traveled 2 meters.
In the second - 1 meter.
In the fifth - 2 meters.
In total, the entire distance traveled is 5 meters.
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (12 points) Find a natural number $n$ such that the numbers $n+30$ and $n-17$ are squares of other numbers.
|
Answer: 546.
Solution. From the condition of the problem, it follows that $\left\{\begin{array}{l}n+30=k^{2} \\ n-17=m^{2}\end{array}\right.$. Subtracting the second equation from the first, we get $k^{2}-m^{2}=47(*)$ or $(k-m)(k+m)=47$. Since 47 is a prime number, the possible cases are $\left\{\begin{array}{l}k-m= \pm 1, \\ k+m= \pm 47\end{array}\right.$ or conversely $\left\{\begin{array}{l}k-m= \pm 47, \\ k+m= \pm 1,\end{array}\right.$ but for any case $k= \pm 24$. Then $n=24^{2}-30=546$. Verification: $n-17=546-17=529=23^{2}$.
Grading criteria. Full solution 12 points. If equation (*) is obtained, 6 points, if all possible cases for the factors are obtained, plus 3 points. If the solution process is correct but there are arithmetic errors, minus 2 points.
|
546
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (15 points) Every day, Ivan Ivanovich is taken to work by a company car. One day, Ivan Ivanovich decided to walk and left the house an hour earlier than usual. On the way, he met the company car and finished the journey in it. As a result, he arrived at work 10 minutes earlier than the usual time. How long did Ivan Ivanovich walk?
|
Answer: 55 minutes.
Solution. Since Ivan Ivanovich saved the driver 10 minutes by walking, the car travels from Ivan Ivanovich's house to the meeting point in 5 minutes.
We get: $u T=5 v$,
(2 points)
where $u$ is Ivan Ivanovich's walking speed, $v$ is the car's speed, and $T$ is the time Ivan Ivanovich walked. The distance from Ivan Ivanovich's house to his workplace: $s=v t$,
and on the other hand: $s=u T+v(t-T+50)$.
Solving the system of equations (1), (2), and (3), we get: $T=55$ minutes.
|
55
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (12 points) Find a natural number $n$ such that the numbers $n+15$ and $n-14$ are squares of other numbers.
|
Answer: 210.
Solution. From the condition of the problem, it follows that $\left\{\begin{array}{l}n+15=k^{2} \\ n-14=m^{2}\end{array}\right.$ Subtracting the second equation from the first, we get $k^{2}-m^{2}=29(*)$ or $(k-m)(k+m)=29$. Since 29 is a prime number, the possible cases are $\left\{\begin{array}{l}k-m= \pm 1 \\ k+m= \pm 29,\end{array}\right.$ or conversely $\left\{\begin{array}{l}k-m= \pm 29, \\ k+m= \pm 1,\end{array}\right.$ but for any case $k= \pm 15$. Then $n=15^{2}-15=210$. Verification: $n-14=210-14=196=14^{2}$.
Grading criteria. Full solution 12 points. If equation (*) is obtained, 6 points, if all possible cases for the factors are obtained, plus 3 points. If the solution process is correct but there are arithmetic errors, minus 2 points.
|
210
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (15 points) Every day, Ivan Ivanovich is taken to work by a company car. One day, Ivan Ivanovich decided to walk and left home one and a half hours earlier than usual. On the way, he met the company car and finished the journey in it. As a result, he arrived at work 20 minutes earlier than the usual time. How long did Ivan Ivanovich walk?
|
Answer: 80 minutes.
Solution. Since Ivan Ivanovich saved the driver 20 minutes by walking, the car takes 10 minutes to drive from Ivan Ivanovich's house to the meeting point.
We get: $u T=10 v$,
(2 points)
where $u$ is Ivan Ivanovich's walking speed, $v$ is the car's speed, and $T$ is the time Ivan Ivanovich walked. The distance from Ivan Ivanovich's house to his workplace: $s=v t$,
and on the other hand: $s=u T+v(t-T+70)$.
(4 points)
Solving the system of equations (1), (2), and (3), we get $T=80$ minutes.
(4 points)
|
80
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Fifteen numbers are arranged in a circle. The sum of any six consecutive numbers is 50. Petya covered one of the numbers with a card. The two numbers adjacent to the card are 7 and 10. What number is under the card?
|
Answer: 8.
Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 15$.) Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{4}=a_{10}=a_{1} .
$$
Now it is clear that for any $i$, $a_{i}=a_{i+3}$, i.e., the numbers go in the following order:
$$
a, b, c, a, b, c, \ldots, a, b, c
$$
From the condition, it follows that
$$
2(a+b+c)=50
$$
Thus, the sum of any three consecutive numbers is 25. Hence the answer.
Scoring. 12 points for a complete solution.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Let $d$ - the greatest common divisor of eight natural numbers, the sum of which is equal to 595. What is the largest value that $d$ can take?
|
Answer: 35.
Solution. If each of the numbers is divisible by $d$, then their sum is also a multiple of $d$. Therefore, $d$ is a divisor of the number 595. Let's factorize the latter into prime factors: $595=5 \cdot 7 \cdot 17$ and list all its divisors:
$$
1,5,7,17,35,85,119,595
$$
Each of the eight numbers (from the problem statement) is not less than $d$. Therefore, their sum, which is 595, is not less than $8 d$. From the inequality $595 \geqslant 8 d$ we get $d \leqslant 74$. From the list $(*)$ the largest possible value is 35.
It is not difficult to come up with a corresponding example (which is far from unique): let seven numbers be 70, and one more be 105.
Evaluation. 13 points for a complete solution. 7 points for an estimate without an example. 6 points for an example without an estimate.
|
35
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (10 points) A rigid board of mass $m$ and length $l=20 m$ is partially lying on the edge of a horizontal surface, hanging off it by three quarters of its length. To prevent the board from falling, a stone of mass $2 m$ is placed at the very edge of the board. How far from the stone can a person of mass $m / 2$ walk along the board? Neglect the sizes of the stone and the person compared to the size of the board.
|
Answer: $15 m$
Solution. The moment rule: $2 m g \cdot \frac{l}{4}=m g \cdot \frac{l}{4}+\frac{m}{2} g \cdot\left(x-\frac{l}{4}\right)$.
As a result, we get $x=\frac{3}{4} l=15$ m.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (15 points) The mass of a vessel completely filled with kerosene is 31 kg. If this vessel is completely filled with water, its mass will be 33 kg. Determine the mass of the empty vessel. The density of water $\rho_{W}=1000 \kappa g / \mathrm{m}^{3}$, the density of kerosene $\rho_{K}=800$ kg $/ \mathrm{m}^{3}$.
|
Answer: 23 kg
Solution. The mass of the vessel filled with kerosene $m_{1}=m_{c}+\rho_{K} V$,
where $m_{c}$ is the mass of the empty vessel, $V$ is the volume occupied by the kerosene. The mass of the vessel filled with water: $m_{2}=m_{c}+\rho_{b} V$. (4 points)
We get that $V=\frac{m_{2}-m_{1}}{\rho_{B}-\rho_{K}}=\frac{33-31}{1000-800}=0.01 \mathrm{~m}^{3}$.
The mass of the empty vessel: $m_{c}=m_{1}-\rho_{K} V=31-800 \cdot 0.01=23$ kg.
## Multidisciplinary Engineering Olympiad "Star"
in natural sciences 7th grade
## Final Stage Variant 2 Problems, answers, and evaluation criteria
$2018-2019$
|
23
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Twenty numbers are arranged in a circle. It is known that the sum of any six consecutive numbers is 24. What is the number in the 12th position if the number in the 1st position is 1?
|
Answer: 7.
Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 20)$. Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{19}=a_{5}=a_{11}=a_{17}=a_{3}=a_{9}=a_{15}=a_{1}
$$
Now it is clear that all numbers at odd positions are equal to each other. The same is true for numbers at even positions. Therefore, the numbers go like this:
$$
x, y, x, y, \ldots, x, y
$$
From the condition, it follows that
$$
3(x+y)=24, \quad x=1
$$
Hence, $y=7$. Therefore, ones are at odd positions, and sevens are at even positions.
Evaluation. 12 points for a complete solution.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Let $d$ be the greatest common divisor of ten natural numbers whose sum is 1001. What is the largest value that $d$ can take?
|
Answer: 91.
Solution. If each of the numbers is divisible by $d$, then their sum is also divisible by $d$. Therefore, $d$ is a divisor of the number 1001. Let's factorize the latter into prime factors: $1001=7 \cdot 11 \cdot 13$ and list all its divisors:
$$
1,7,11,13,77,91,143,1001
$$
Each of the ten numbers (from the problem statement) is not less than $d$. Therefore, their sum, which is 1001, is not less than $10d$. From the inequality $1001 \geqslant 10d$ we get $d \leqslant 101$. From the list $(*)$ the largest possible value is 91.
It is not difficult to come up with a corresponding example (it is unique): nine numbers are equal to 91, and one more is 182.
Scoring. 13 points for a complete solution. 7 points for an estimate without an example. 6 points for an example without an estimate.
|
91
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (10 points) A rigid board of mass $m$ and length $l=24$ m is partially lying on the edge of a horizontal surface, hanging off it by two-thirds of its length. To prevent the board from falling, a stone of mass $2 m$ is placed at the very edge of the board. How far from the stone can a person of mass $m$ walk along the board? Neglect the sizes of the stone and the person compared to the size of the board.
|
Answer: $20 m$
Solution. The moment rule: $2 m g \cdot \frac{l}{3}=m g \cdot \frac{l}{6}+m g \cdot\left(x-\frac{l}{3}\right)$.
As a result, we get $x=\frac{5}{6} l=20 m$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (15 points) The mass of a vessel that is completely filled with kerosene is 20 kg. If this vessel is completely filled with water, its mass will be 24 kg. Determine the mass of the empty vessel. The density of water $\rho_{W}=1000 \kappa g / \mu^{3}$, the density of kerosene $\rho_{K}=800 \kappa g / \mathrm{M}^{3}$.
|
Answer: 4 kg
Solution. The mass of the vessel filled with kerosene: $m_{1}=m_{c}+\rho_{K} V$, where $m_{c}$ - the mass of the empty vessel, $V$ - the volume occupied by the kerosene.
The mass of the vessel filled with water $m_{2}=m_{c}+\rho_{B} V$.
(4 points)
We get that $V=\frac{m_{2}-m_{1}}{\rho_{B}-\rho_{K}}=\frac{24-20}{1000-800}=0.02 \mu^{3}$.
The mass of the empty vessel: $m_{c}=m_{1}-\rho_{\kappa} V=20-800 \cdot 0.02=4$ kg.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (12 points) There are four weights of different masses. Katya weighs the weights in pairs. As a result, she got 1800, 1970, 2110, 2330, and 2500 grams. How many grams does the sixth weighing variant weigh?
|
Answer: 2190.
Solution. Let $x, y, z, t$ be the weight of each weight. Then the pairwise weighings will be $x+y, x+z, x+t, y+z, y+t, z+t$ grams. There are three pairs of given numbers: 1) $x+y$ and $z+t$, 2) $x+z$ and $y+t$, 3) $x+t$ and $y+z$, with the same total weight $x+y+z+t$. Let's find pairs with the same sum: $1800+$ $2500=1970+2330=4300$. Notice that there are no other pairs of given numbers with the same sum. Then the sixth weighing option: $4300-2110=$ 2190.
Grading criteria. Full solution - 12 points. If the participant noticed that three pairs out of six numbers - the total weights of two weights, when summed, give the weight of all weights, and therefore must be the same, give 7 points. The total weight of all weights is found - another 2 points. Noted that there are no other pairs of given numbers with the same sum, except for the found ones - another 2 points. Arithmetic errors in the correct solution process - minus 2 points.
|
2190
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (13 points) Sixteen people are standing in a circle: each of them is either a truth-teller (he always tells the truth) or a liar (he always lies). Everyone said that both of their neighbors are liars. What is the maximum number of liars that can be in this circle?
|
Answer: 10.
Solution. A truthful person can only be next to liars. Three liars in a row cannot stand, so between any two nearest truthful persons, there is one or two liars. Then, if there are 5 or fewer truthful persons, in the intervals between them, there can be no more than 10 liars in total, so there are no more than 15 people in total. This leads to a contradiction. Therefore, there are no fewer than 6 truthful persons, and no more than 10 liars. An example with 10 liars is easily provided.
Evaluation Criteria. Full solution - 13 points. Example constructed - 3 points. Estimate proven - 9 points. Correct estimate named but not justified by considering examples - 2 points.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (12 points) There are four weights of different masses. Katya weighs the weights in pairs. As a result, she got 1700, 1870, 2110, 2330, and 2500 grams. How many grams does the sixth weighing variant weigh?
|
Answer: 2090.
Solution. Let $x, y, z, t$ be the weight of each weight. Then the pairwise weighings will be $x+y, x+z, x+t, y+z, y+t, z+t$ grams. There are three pairs: 1) $x+$ y and $z+t, 2) x+z$ and 3) $y+t, x+t$ and $y+z$ with the same weight $x+y+z+t$. Let's find pairs with the same sum: $1700+2500=$
$1870+2330$. Notice that there are no other pairs of these numbers with the same sum. Then the sixth weighing option: $4200-2110=2090$.
Grading criteria. Full solution - 12 points. If the participant noticed that three pairs out of six numbers - the total weights of two weights, when summed, give the weight of all weights, and therefore must be the same, give 7 points. The total weight of all weights is found - another 2 points. Noted that there are no other pairs of these numbers with the same sum, except for the found ones - another 2 points. Arithmetic errors in the correct solution process - minus 2 points.
|
2090
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (13 points) In a circle, there are 17 people: each of them is either a truth-teller (he always tells the truth) or a liar (he always lies). Everyone said that both of their neighbors are liars. What is the maximum number of liars that can be in this circle?
|
Answer: 11.
Solution. A truth-teller can only be surrounded by liars. Three liars in a row cannot stand, so between any two nearest truth-tellers, there is one or two liars. Then, if there are 5 or fewer truth-tellers, there can be no more than 10 liars in the gaps between them, making a total of no more than 15 people. This leads to a contradiction. Therefore, there must be at least 6 truth-tellers, and no more than 12 liars, but there are 17 people in total, so there must be 11 liars. An example with 11 liars is easily provided.
Grading Criteria. Full solution - 13 points. Example constructed - 3 points. Estimate proven - 9 points. Correct estimate named but not justified by considering examples - 2 points.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The sofa cost 62500 rubles. Once a month, its price changed by $20 \%$ either increasing or decreasing. It is known that over the course of six months, the price increased three times and decreased three times (the order in which this happened is unknown). Can the final price of the sofa after six months be determined uniquely? If so, what did it become?
|
Answer: Yes; the sofa will cost 55296 rubles.
Solution. An increase in price by $20 \%$ means the current price is multiplied by $6 / 5$, and a decrease in price by $20 \%$ means the current price is multiplied by $4 / 5$. Therefore, regardless of the order in which the price increased or decreased, the price of the sofa after six months will be $62500 \cdot(6 / 5)^{3} \cdot(4 / 5)^{3}=$ 55296 rubles.
Evaluation. If only some specific cases are considered and the correct numerical answer is obtained, 4 points. For a complete solution, 12 points.
|
55296
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. How many five-digit numbers exist where the sum of the first two digits is half the sum of the last two digits?
|
Answer: 2250.
Solution. Let the first two digits of the number be $a$ and $b$, and the last two digits be $c$ and $d$. We need to find the number of combinations of these digits for which $2(a+b)=c+d$. Since $c+d \leqslant 18$, then $a+b \leqslant 9$.
Let $x_{i}(i=1,2, \ldots, 9)$ denote the number of pairs of the first two digits whose sum is $i$. The first digit can take values from 1 to $i$, after which the second digit is uniquely determined. Thus, $x_{i}=i$.
Let $y_{i}(i=1,2, \ldots, 9)$ denote the number of pairs of the last two digits whose sum is $2i$. Each time, the digit $d$ is uniquely determined by the digit $c$: $d=2i-c$. Therefore, we need to count the number of possible values for the digit $c$. If $i \leqslant 4$, i.e., $2i \leqslant 8$, then the digit $c$ can take values from 0 to $2i-$ a total of $2i+1$ values. In other words, in this case, $y_{i}=2i+1$. If $5 \leqslant i \leqslant 9$, i.e., $10 \leqslant 2i \leqslant 18$, then the digit $c$ can take values from $2i-9$ to $9-$ a total of $19-2i$ values, and $y_{i}=19-2i$.
Let's compile the found values of $x_{i}$ and $y_{i}$ into one table.
| $i$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x_{i}$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| $y_{i}$ | 3 | 5 | 7 | 9 | 9 | 7 | 5 | 3 | 1 |
The number of five-digit numbers in which $a+b=i$ and $c+d=2i$ is $10 x_{i} y_{i}$ (the middle digit can be chosen in 10 ways). Therefore, the total number of the desired numbers is $10\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{9} y_{9}\right)$. Substituting the found values gives the answer 2250.
Evaluation. 13 points for a complete solution. If all $x_{i}$ and $y_{i}$ are correctly found but there is an arithmetic error in the final calculation, 11 points. If the solution approach is correct but some $x_{i}$ or $y_{i}$ are incorrectly determined, 4-5 points.
|
2250
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) Two wheels rotate, meshed with each other, around fixed axes passing through the centers of wheels $A$ and $B$. The radii of the wheels differ by a factor of three. The smaller wheel makes 30 revolutions per minute. Determine how many seconds the larger wheel spends on one revolution?
|
Answer: $6 s$
Solution. The speeds of points lying on the edge of the wheels are the same.
(2 points)
The distances traveled by these points differ by three times.
The small wheel spends on one revolution: $t_{\text {small }}=\frac{1 \text { min }}{30 \text { revolutions }}=\frac{60 s}{30 \text { revolutions }}=2 s$. (3 points)
Therefore, the large wheel spends on one revolution:
$t_{\text {large }}=3 t_{\text {small }}=3 \cdot 2=6 s$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (15 points) Density is defined as the ratio of the mass of a body to its volume. There are two cubes. The second cube is made of a material with twice the density of the first, and the side length of the second cube is 100% greater than the side length of the first. By what percentage is the mass of the second cube greater than the mass of the first?
|
Answer: by $1500 \%$
Solution. From the condition, we get $\rho_{2}=2 \rho_{1}$, i.e., $\frac{m_{2}}{V_{2}}=2 \cdot \frac{m_{1}}{V_{1}}$.
Also, from the condition, it follows that the side length of the second cube $a_{2}=2 a_{1}$.
(2 points)
Therefore, their volumes are related by the ratio $V_{2}=a_{2}^{3}=\left(2 a_{1}\right)^{3}=8 V_{1}$.
(3 points)
We obtain $\frac{m_{2}}{8 V_{1}}=2 \cdot \frac{m_{1}}{V_{1}}$. (2 points) From which it follows that $m_{2}=16 m_{1}$.
That is, the mass of the second cube is greater by $1500 \%$.
Multidisciplinary Engineering Olympiad "Star" in Natural Sciences
6th grade
Final Stage
$2018-2019$
Variant 2
Tasks, answers, and evaluation criteria
|
1500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The item cost 64 rubles. Once a month, its price changed by $50\%$ either increasing or decreasing. It is known that over the course of six months, the price increased three times and decreased three times (the order in which this happened is unknown). Can the final price of the item after six months be determined uniquely? If so, what did it become?
|
Answer: Yes; the item will cost 27 rubles.
Solution. An increase of $50\%$ means that the current price is multiplied by $3/2$, and a decrease of $50\%$ means that the current price is multiplied by $1/2$. Therefore, regardless of the order in which the price increased or decreased, the price of the sofa after six months will be $64 \cdot (3/2)^3 \cdot (1/2)^3 = 27$ rubles.
Evaluation. If only some specific cases are considered and the correct numerical answer is obtained, 4 points. For a complete solution, 12 points.
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. How many five-digit numbers exist where the sum of the first two digits is twice the sum of the last two digits?
|
Answer: 2600.
Solution. Let the first two digits of the number be $a$ and $b$, and the last two digits be $c$ and $d$. We need to find the number of combinations of these digits for which $a+b=2(c+d)$. Since $a+b \leqslant 18$, then $c+d \leqslant 9$.
Let $x_{i}(i=1,2, \ldots, 9)$ denote the number of pairs of the last two digits whose sum is $i$. The first digit can take values from 0 to $i$, after which the second digit is uniquely determined. Thus, $x_{i}=i+1$.
Let $y_{i}(i=1,2, \ldots, 9)$ denote the number of pairs of the first two digits whose sum is $2i$. Each time, the digit $b$ is uniquely determined by the digit $a$: $b=2i-a$. Therefore, we need to count the number of possible values for the digit $a$. If $i \leqslant 4$, i.e., $2i \leqslant 8$, then the digit $a$ can take values from 1 to $2i$ - a total of $2i$ values. In other words, in this case, $y_{i}=2i$. If $5 \leqslant i \leqslant 9$, i.e., $10 \leqslant 2i \leqslant 18$, then the digit $a$ can take values from $2i-9$ to $9$ - a total of $19-2i$ values, and $y_{i}=19-2i$.
Let's compile the found values of $x_{i}$ and $y_{i}$ into one table.
| $i$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x_{i}$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| $y_{i}$ | 2 | 4 | 6 | 8 | 9 | 7 | 5 | 3 | 1 |
The number of five-digit numbers in which $c+d=i$ and $a+b=2i$ is $10 x_{i} y_{i}$ (the middle digit can be chosen in 10 ways). Therefore, the total number of the desired numbers is $10\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{9} y_{9}\right)$. Substituting the found values gives the answer 2600.
Evaluation. 13 points for a complete solution. If all $x_{i}$ and $y_{i}$ are correctly found but there is an arithmetic error in the final calculation, 11 points. If the solution approach is correct but some $x_{i}$ or $y_{i}$ are incorrectly determined, 4-5 points.
|
2600
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (10 points) An Englishman was the owner of a plot of land in Russia. He knows that, in the units familiar to him, the size of his plot is three acres. The cost of the land is 250000 rubles per hectare. It is known that 1 acre $=4840$ square yards, 1 yard $=0.9144$ meters, 1 hectare $=10000 m^{2}$. Calculate how much the Englishman will receive as a result of the sale.
|
Answer: approximately 303514 rubles
Solution. 1 square yard $=0.9144 \cdot 0.9144=0.83612736 \mu^{2}$,
3 acres $=3 \cdot 4840=14520$ square yards $a=14520 \cdot 0.83612739=12140.57 \mu^{2}$ (2 points), 12140.57 $\mu^{2}=\frac{12140.57 \mu^{2}}{10000} \approx 1.214057 ~ 2 a$,
$$
1.214057 \cdot 250000=303514.25 \text { rubles }
$$
|
303514
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) Two wheels rotate, meshed with each other, around fixed axes passing through the centers of wheels $A$ and $B$. The radii of the wheels differ by a factor of three. The larger wheel makes 10 revolutions per minute. Determine how many seconds the smaller wheel spends on one revolution?
|
Answer: $2 c$
Solution. The speeds of points lying on the edge of the wheels are the same.
The distances traveled by these points differ by three times.
The large wheel spends on one revolution
$$
t_{\text {large }}=\frac{1 \text { min }}{10 \text { revolutions }}=\frac{60 \text { s }}{10 \text { revolutions }}=6 \text { s }
$$
Therefore, the small wheel spends on one revolution $t_{\text {small }}=\frac{1}{3} t_{\text {large }}=\frac{6}{3}=2 c$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (15 points) Density is defined as the ratio of the mass of a body to its volume. There are two cubes. The second cube is made of a material with half the density of the first, and the side length of the second cube is 100% greater than the side length of the first. By what percentage is the mass of the second cube greater than the mass of the first?
|
Answer: by $300 \%$
Solution. From the condition, it follows that $\rho_{2}=\frac{1}{2} \rho_{1}$, i.e., $\frac{m_{2}}{V_{2}}=\frac{1}{2} \cdot \frac{m_{1}}{V_{1}}$.
It also follows from the condition that the side length of the second cube $a_{2}=2 a_{1}$.
(2 points)
Therefore, their volumes are related by the ratio $V_{2}=a_{2}^{3}=\left(2 a_{1}\right)^{3}=8 V_{1}$.
(3 points)
We obtain $\frac{m_{2}}{8 V_{1}}=\frac{1}{2} \cdot \frac{m_{1}}{V_{1}}$. (2 points) From which it follows that $m_{2}=4 m_{1}$.
That is, the mass of the second cube is greater by $300 \%$.
|
300
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (12 points) The sequence of functions is defined by the formulas
$$
f_{0}(x)=3 \sin x, f_{n+1}(x)=\frac{9}{3-f_{n}(x)}
$$
for any integer $n \geq 0$. Find $f_{2023}\left(\frac{\pi}{6}\right)$.
|
Answer: $f_{2023}\left(\frac{\pi}{6}\right)=6$.
Solution. It is easy to compute: $f_{3}(x)=f_{0}(x)$, therefore $f_{2023}(x)=f_{1}(x)=$ $\frac{9}{3-3 \sin x}$. Consequently, $f_{2023}\left(\frac{\pi}{6}\right)=6$.
Remark. One can immediately compute the values of the functions at the given point. This results in a cyclic sequence
$$
f_{0}\left(\frac{\pi}{6}\right)=\frac{3}{2} ; f_{1}\left(\frac{\pi}{6}\right)=6 ; f_{2}\left(\frac{\pi}{6}\right)=-3 ; f_{3}\left(\frac{\pi}{6}\right)=\frac{3}{2} ; \ldots
$$
Grading criteria. Full solution - 12 points. The relation $f_{3}(x)=f_{0}(x)$ is found - 7 points, the equality $f_{2023}(x)=f_{1}(x)$ is found - another 4 points. Calculation errors - minus 1 point.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (12 points) The sequence of functions is defined by the formulas
$$
f_{0}(x)=2 \cos x, f_{n+1}(x)=\frac{4}{2-f_{n}(x)}
$$
for any integer $n \geq 0$. Find $f_{2023}\left(\frac{\pi}{3}\right)$.
|
Answer: $f_{2023}\left(\frac{\pi}{3}\right)=4$.
Solution. It is easy to compute: $f_{3}(x)=f_{0}(x)$, therefore $f_{2023}(x)=f_{1}(x)=$ $\frac{4}{2-2 \cos x}$. Consequently, $f_{2023}\left(\frac{\pi}{3}\right)=4$.
Remark. One can immediately compute the values of the functions at the given point. This results in a cyclic sequence
$$
f_{0}\left(\frac{\pi}{3}\right)=1 ; f_{1}\left(\frac{\pi}{3}\right)=4 ; f_{2}\left(\frac{\pi}{3}\right)=-2 ; f_{3}\left(\frac{\pi}{3}\right)=1 ; \ldots
$$
Grading criteria. Full solution - 12 points. The relation $f_{3}(x)=f_{0}(x)$ is found - 7 points, the equality $f_{2023}(x)=f_{1}(x)$ is found - another 4 points. Calculation errors - minus 1 point.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. $\left(10\right.$ points) An arc, with a central angle of $\alpha=60^{\circ}$, is cut from a circle with radius $R=40 \mathrm{~cm}$. A charge $q=5$ μC is uniformly distributed along the arc. Determine the electric field strength $E$ at the center of curvature of this arc.
|
Answer: 269 kV/m.
Solution. Consider a small element of the arc of length $d l$, on which charge $d q$ is located. It creates a field strength at the point of interest:
$d E=k \frac{d q}{R^{2}}$.
The projection of this field strength onto the horizontal direction is:
$d E \cos \beta=k \frac{d q}{R^{2}} \cos \beta$.
The charge of this element is: $d q=\frac{q}{R \cdot \frac{\pi}{3}} \cdot d l$.
We obtain: $d E \cos \beta=k \frac{\frac{q}{R \cdot \frac{\pi}{3}} \cdot d l}{R^{2}} \cos \beta=k \frac{3 q \cdot d l \cdot \cos \beta}{\pi R^{3}}$.
Note that: $d l \cdot \cos \beta=d y$.
Summing over all such elements, we get:
$E=k \frac{3 q \cdot 2 R \sin 30^{\circ}}{\pi R^{3}}=6 k \frac{q \sin 30^{\circ}}{\pi R^{2}}=6 \cdot 9 \cdot 10^{9} \frac{5 \cdot 10^{-6} \sin 30^{\circ}}{\pi \cdot 0.4^{2}}=268574 \frac{\mathrm{V}}{\mathrm{m}}$.
|
269
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (16 points) Solve the equation $x-5=\frac{3 \cdot|x-2|}{x-2}$. If the equation has multiple roots, write their sum in the answer.
#
|
# Answer: 8.
Solution. The equation has a restriction on the variable $x \neq 2$. We open the modulus: for $x>2, x-5=3, x=8$. For $x<2, \quad x-5=-3, x=2$ - an extraneous root.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) Given a rectangle $A B C D$. On two sides of the rectangle, different points are chosen, five points on $A B$ and six on $B C$. How many different triangles exist with vertices at the chosen points?
|
Answer: 135.
Solution. To form a triangle, one needs to choose two points on one side and one point on another. We have: 5 ways to choose the first point on $AB$, 4 ways - the second, and since the triangle does not change with the permutation of its vertices, we divide $5 \cdot 4$ by 2. Thus, $\frac{5 \cdot 4}{2}=10$ ways to choose two points on side $AB$, and 6 ways on side $BC$. In total, $10 \cdot 6=60$ ways. Similarly, $\frac{6 \cdot 5}{2}=15$ ways to choose two points on side $BC$ and 5 ways to choose one point on side $AB$. In total, $15 \cdot 5=75$ ways. Therefore, there are 135 different triangles with vertices at the chosen points.
|
135
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (17 points) In triangle $A B C$, a point $N$ is taken on side $A C$ such that $A N=N C$. Side $A B$ is twice as long as $B N$, and the angle between $A B$ and $B N$ is $50^{\circ}$. Find the angle $A B C$.
|
Answer: 115.
Solution. Complete triangle $A B C$ to parallelogram $A B C D$.
Then, $D B=2 N B=A B$. Therefore, triangle $A B D$ is isosceles and $\angle A D B=\frac{180^{\circ}-50^{\circ}}{2}=65^{\circ}$. $\angle A D B=\angle D B A$ (alternate interior angles when lines are parallel). In the end, we get $\angle A B C=\angle A B D+\angle D B C=50^{\circ}+65^{\circ}=115^{\circ}$.
|
115
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (20 points) Determine the angle between the hour and minute hands at the moment when they show 13 hours and 20 minutes.
|
Answer: $80^{\circ}$.
Solution. The minute hand has moved away from twelve by $\frac{20}{60} \cdot 360=120^{0}$. The hour hand has moved away from twelve by $\frac{1}{12} \cdot 360+\frac{20}{60} \cdot \frac{1}{12} \cdot 360=40^{\circ}$. The angle between the hands $120^{\circ}-40^{\circ}=80^{\circ}$.
|
80
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (16 points) Solve the equation $x-7=\frac{4 \cdot|x-3|}{x-3}$. If the equation has multiple roots, write their sum in the answer.
|
Answer: 11.
Solution. The equation has a restriction on the variable $x \neq 3$. We open the modulus: for $x>3, x-7=4, x=11$. For $x<3, \quad x-7=-4, x=3-$ extraneous root.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) Given a rectangle $A B C D$. On two sides of the rectangle, different points are chosen, six points on $A B$ and seven - on $B C$. How many different triangles exist with vertices at the chosen points?
|
Answer: 231.
Solution. To form a triangle, one needs to choose two points on one side and one point on another. There are 6 ways to choose the first point on $AB$, 5 ways to choose the second, and since the triangle does not change with the permutation of its vertices, we divide $6 \cdot 5$ by 2. Thus, $\frac{6 \cdot 5}{2}=15$ ways to choose two points on side $AB$, and 7 ways on side $BC$. In total, $15 \cdot 7=105$ ways. Similarly, $\frac{7 \cdot 6}{2}=21$ ways to choose two points on side $BC$ and 6 ways to choose one point on side $AB$. In total, $21 \cdot 6=126$ ways. Therefore, there are 231 different triangles with vertices in the chosen points.
|
231
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (17 points) In triangle $A B C$, a point $N$ is taken on side $A C$ such that $A N=N C$. Side $A B$ is twice as long as $B N$, and the angle between $A B$ and $B N$ is $40^{\circ}$. Find the angle $A B C$.
|
Answer: 110.
## Solution.
Extend triangle $ABC$ to parallelogram $ABCD$. Then, $DB=2NB=AB$. Therefore, triangle $ABD$ is isosceles and $\angle ADB=\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ}$. $\angle ADB=\angle DBA$ (alternate interior angles when lines are parallel). In the end, we get $\angle ABC=\angle ABD+\angle DBC=40^{\circ}+70^{\circ}=110^{\circ}$.
|
110
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (20 points) Determine the angle between the hour and minute hands at the moment when they show 15 hours and 40 minutes.
|
Answer: $130^{\circ}$.
Solution. The minute hand has moved away from twelve by $\frac{40}{60} \cdot 360=240^{\circ}$. The hour hand has moved away from twelve by $\frac{3}{12} \cdot 360+\frac{40}{60} \cdot \frac{1}{12} \cdot 360=110^{\circ}$. The angle between the hands: $240^{\circ}-110^{\circ}=130^{\circ}$.
|
130
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The cold water tap fills the bathtub in 17 minutes, and the hot water tap in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water in it?
|
Answer: in 3 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 8.5 minutes. Therefore, the hot water tap should be open for 3 minutes longer.
Evaluation. 12 points for the correct solution.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A certain mechanism consists of 30 parts, some of which are large, and some are small. It is known that among any 12 parts taken, there will be at least one small part, and among any 20 parts - at least one large part. How many of each type of part does the mechanism contain?
|
Answer: 11 large parts and 19 small parts.
Solution. Since among any 12 parts there is a small one, there are no more than 11 large parts. Since among any 20 parts there is a large one, there are no more than 19 small parts. If there were fewer than 11 large parts or fewer than 19 small parts, there would be fewer than 30 parts in total, but according to the condition, there are 30. Therefore, there are 11 large parts and 19 small parts.
Grading. 13 points for a correct solution. 2 points for a correct answer (without justification). If the correctness of the answer is shown but its uniqueness is not justified, 6 points.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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